The limit of $(1+x^2+y^2)^{1\over x^2 + y^2 +xy^2}$ as $(x,y)$ approaches $(0.0)$ Im having trouble solving the following limit problem:
Whats the limit of $(1+x^2+y^2)^{1\over x^2 + y^2 +xy^2}$ as $(x,y)$ approaches $(0.0)$. 
I know that the first step is to change $(x,y)$ to polar coordinates. However after I've done that and simplified the expression I'm left with the following: 
$$
(1+r^2)^{1\over r^2+r^2\cos(α) \sin^2(α)}. 
$$
How do i prove that the expression goes to $e$ when $r→0$?
Thanks in advance!
 A: You can write it using the limit laws, esp. the one on composition of continuous functions, as
$$
\lim_{(x,y)\to (0,0)}(1+x^2+y^2)^{1\over x^2+y^2+xy^2}=\lim_{(x,y)\to (0,0)}\left(\lim_{(x,y)\to (0,0)}(1+x^2+y^2)^{1\over x^2+y^2}\right)^{x^2+y^2 \over x^2+y^2+xy^2}
\\
=\lim_{(x,y)\to (0,0)} \exp\left({x^2+y^2 \over x^2+y^2+xy^2}\right)
$$
etc.
A: Let $\epsilon>0$. Then for $|x|<\epsilon$:
$$
(1+x^2+y^2)^{\frac{1}{x^2+y^2+\epsilon y^2+\epsilon x^2}} \leq
(1+x^2+y^2)^{\frac{1}{x^2+y^2+x y^2}} \leq
(1+x^2+y^2)^{\frac{1}{x^2+y^2-\epsilon y^2-\epsilon y^x}}.
$$
Since 
$$
(1+x^2+y^2)^{\frac{1}{x^2+y^2-\epsilon y^2-\epsilon y^x}}=
\Big((1+x^2+y^2)^{\frac{1}{x^2+y^2}}\Big)^{\frac{1}{(1-\epsilon)}} 
\rightarrow e^{\frac{1}{(1-\epsilon)}} \text{ as }(x,y)\rightarrow 0
$$
and
$$
(1+x^2+y^2)^{\frac{1}{x^2+y^2+\epsilon y^2+\epsilon y^x}}=
\Big((1+x^2+y^2)^{\frac{1}{x^2+y^2}}\Big)^{\frac{1}{(1+\epsilon)}} 
\rightarrow e^{\frac{1}{(1+\epsilon)}} \text{ as }(x,y)\rightarrow 0
$$
it follows that 
$$ e^{\frac{1}{(1+\epsilon)}} \leq \lim_{(x,y)\rightarrow 0} (1+x^2+y^2)^{\frac{1}{x^2+y^2+x y^2}} \leq e^{\frac{1}{(1-\epsilon)}}.$$
The above is true for all $\epsilon>0$. Therefore we must have 
$$ \lim_{(x,y)\rightarrow 0} (1+x^2+y^2)^{\frac{1}{x^2+y^2+x y^2}} = e^1 = e.$$ 
